Radicals test in Year 2 of high school

If there is one chapter that in Year 2 makes the grades of students who were doing fine until October collapse, it is radicals. The test almost always falls between the end of November and the days before the Christmas holidays, and in my experience it is the one that leads the most parents to write to me in December: “he has always had 7, he got 4 and we don’t understand why”.

The reason is that radicals look easy. The rules are five or six, the book sets them out in three pages, and a student who has understood powers thinks they are fine. Then the test arrives, and two things happen: the conditions of existence are forgotten, and the properties are applied even where they don’t hold. It is not a problem of understanding, it is a problem of precision — which is another thing, and is trained differently.

Here you’ll find what the test really asks, the method for each type of exercise, three exercises worked step by step and the four mistakes I see repeated in every class.

The basic idea: a radical is a power in disguise

Before the rules, a sentence to keep in mind for the whole chapter:

n a^{m} = a^{\frac{m}{n}}

A radical is a power with a fractional exponent. It is not a teaching analogy: it is the definition the book gives (almost always at the end of the chapter, which is a pity, because put at the start it would explain everything else).

This equality has an enormous practical consequence: all the properties of radicals are the properties of powers, rewritten. The invariance property is the simplification of a fraction in the exponent. The product of two roots with the same index is the rule $a^{x} \cdot a^{y} = a^{x + y}$ . The root of a root is $(a^{x})^{y} = a^{x y}$ .

If you find yourself memorising six unconnected rules, you are studying the chapter in the way that costs the most. It is worth rewriting each property in both notations, side by side, the first evening you work on it.

Conditions of existence: where marks are really lost

The delicate point is that the equality above holds only for $a \geq 0$ when the index is even. And this is exactly where the test is waiting for you.

The operating rule is short:

even index (, $4$ , $6$ ): the radicand must be greater than or equal to zero;
odd index ( $3$ , $5$ ): no condition, the radicand can be negative.

So $3 - 8 = - 2$ exists and is perfectly legitimate, while $- 8$ in $R$ doesn’t exist.

The second half of the rule is the one most often forgotten, and it is worth writing out in full because it is the number one source of mistakes in tests:

n a^{n} = ∣ a ∣ if n is even, n a^{n} = a if n is odd

In words: when you bring out of the root a factor raised to an even exponent and the index is even, it comes out with the absolute value, not bare. $x^{2} = ∣ x ∣$ , not $x$ . If the teacher has put in an exercise with a letter under the root, they are almost certainly checking this.

The invariance property and simplification

The invariance property says you can multiply (or divide) the index and the exponent by the same number, as long as the radicand is not negative:

n a^{m} = n \cdot p a^{m \cdot p} (a \geq 0, p \in N, p \neq = 0)

It serves two purposes, and both come up in the test.

Simplifying. $6 a^{4} = 3 a^{2}$ : I divided index and exponent by 2. In fractional-exponent notation it is trivial, $a^{4/6} = a^{2/3}$ , and it is the right way to check it if you have a doubt.

Reducing to the same index, which is needed to multiply or compare different radicals. To multiply $2 \cdot 35$ you bring both to index 6, that is to the least common multiple of the indices:

2 \cdot 35 = 6 2^{3} \cdot 6 5^{2} = 6 8 \cdot 25 = 6200

Mind the condition $a \geq 0$ : it is not a formal detail. $4 (- 3)^{2}$ equals $3$ , whereas applying the property literally would give $- 3$ , which doesn’t exist. With letters you always write the absolute value: $4 a^{2} = ∣ a ∣$ .

Operations: what you can do and what you can’t

This is the table worth keeping in front of you while you do exercises.

Operation	Allowed?	Formula
Product, same index	yes	$n a \cdot n b = n ab$
Quotient, same index	yes	$\frac{n a}{n b} = n \frac{a}{b}$
Product, different indices	yes, after reducing to the same index	LCM of the indices
Power of a radical	yes	$(n a)^{m} = n a^{m}$
Root of a root	yes	$m n a = mn a$
Sum	no	$a + b \neq = a + b$

The last row is the most important rule of the chapter. $9 + 16 = 3 + 4 = 7$ , while $9 + 16 = 25 = 5$ . They are different numbers, and thirty seconds with a calculator are enough to convince yourself for ever.

Sums are done only between like radicals, that is with the same index and the same radicand after bringing out everything that can be brought out. It is exactly the mechanism of like monomials from last year: $32 + 52 = 82$ works like $3 x + 5 x = 8 x$ .

Bringing in and out of the radical sign

Out you bring a factor that appears in the radicand with an exponent greater than or equal to the index:

72 = 36 \cdot 2 = 62

The invisible step is the prime factorisation: $72 = 2^{3} \cdot 3^{2}$ , so a 2 and a 3 come out, that is 6. If you get stuck on bringing out, the problem is almost always not radicals but factorisation — which is the chapter before.

In you bring a coefficient by raising it to the index, and here there is the sign trap:

32 = 3^{2} \cdot 2 = 18, - 32 = - 18

A negative coefficient stays outside. Writing $- 32 = 18$ is a mistake worth half an exercise, because it changes the sign of the result.

Rationalising: three cases, three moves

Rationalising means removing the root from the denominator. The cases that come up in the test are three and they are recognisable at a glance.

Denominator	I multiply numerator and denominator by	Example
$a$	$a$	$\frac{6}{3} = \frac{6 3}{3} = 23$
$n a^{m}$	$n a^{n - m}$	$\frac{1}{5 a ^{2}} = \frac{5 a ^{3}}{a}$
$a \pm b$	the conjugate binomial $a \mp b$	see exercise 2

The third case is the one that almost always makes the “hard” exercise of the test, but it is mechanical: you multiply by the conjugate, and in the denominator you apply the sum times difference $(x + y) (x - y) = x^{2} - y^{2}$ , which makes both roots vanish in one go.

Three worked exercises

1. Sum of unlike radicals (that become like)

50 - 318 + 8

None of the three is like the others as it stands. You bring out everything possible by factorising the radicands:

50188 = 5^{2} \cdot 2 = 52 = 3^{2} \cdot 2 = 32 = 2^{2} \cdot 2 = 22

Now they are all like and you add the coefficients:

52 - 3 \cdot 32 + 22 = 52 - 92 + 22 = - 22

The $3$ in front of $18$ has to be multiplied by the $3$ that came out of the root: forgetting it is the most frequent mistake in this type of exercise.

2. Rationalisation with the conjugate

\frac{2}{5 - 3}

The denominator is a binomial: you multiply above and below by the conjugate $5 + 3$ .

\frac{2}{5 - 3} \cdot \frac{5 + 3}{5 + 3} = \frac{2 ( 5 + 3 )}{( 5 ) ^{2} - ( 3 ) ^{2}}

In the denominator what remains is $5 - 3 = 2$ , and it cancels with the 2 in the numerator:

\frac{2 ( 5 + 3 )}{2} = 5 + 3

A clean result like this is no accident: in tests the numbers are almost always chosen so that the denominator simplifies. If you get a monster, recheck the conjugate — it is easy to get the sign wrong.

3. The radical with a letter (the exercise that separates the grades)

x^{2} - 6 x + 9

First you factorise the radicand, which is a square of a binomial:

x^{2} - 6 x + 9 = (x - 3)^{2}

Then you apply the even-index rule, with the absolute value:

(x - 3)^{2} = ∣ x - 3∣

And the absolute value is spelled out: it equals $x - 3$ if $x \geq 3$ , and $3 - x$ if $x < 3$ . Writing $x - 3$ directly is the mistake the teacher is looking for: for $x = 1$ it would give $- 2$ , that is a negative square root.

A note on the conditions of existence of this exercise: $(x - 3)^{2}$ is a square, so it is always $\geq 0$ and the radical exists for every real $x$ . But the condition of existence must be written all the same, because it is what shows you have checked it.

The four mistakes that cost the most marks

1. Adding under the root. $a + b \neq = a + b$ , always. It is the most serious mistake because it is conceptual: whoever makes it hasn’t yet connected radicals and powers.

2. Bringing out without the absolute value. $x^{2} = ∣ x ∣$ . With numbers you never notice, because they are positive; with letters it costs the exercise.

3. Forgetting the conditions of existence on fractional radicals. If there is a fraction under the root, two conditions are needed together: the radicand $\geq 0$ and the denominator $\neq = 0$ . You lose half a mark at a time, but on a test of six exercises it makes a whole grade.

4. Bringing in a negative coefficient. The minus sign stays outside the root with an even index. Always.

There is then a fifth mistake which isn’t about radicals but sinks them all the same: slow factorisation. If it takes you half a minute to factorise $72$ or to recognise $x^{2} - 6 x + 9$ , you won’t finish the test. In that case the priority is not to review radicals but to redo twenty factorisations from Year 1, and it is time much better spent.

The two weeks before the test

Write the properties on a blank sheet, every day for a week. Two minutes. Next to each, the fractional-exponent version. The one that doesn’t come out is your priority for the day.
Do exercises divided by type, not at random. Five simplifications in a row, then five sums, then five rationalisations. The test requires you to recognise the type in ten seconds, and recognition is trained by seeing similar exercises close together, not alternated.
In the last two days, a timed paper in the format of the test, marked by classifying each mistake into three categories: rule not known, wrong calculation, condition of existence forgotten. They are three different problems with three different remedies, and telling them apart is worth more than another ten exercises.

If the difficulty is not preparation but the blank of the first five minutes, I wrote about it in how to manage test anxiety in maths and physics.

How we work on this in lessons

With radicals my order is always the same: first the rewriting as a power, then the conditions of existence, and only after that the calculation. Whoever starts from the calculation learns to do exercises that resemble those seen; whoever starts from the definition also recognises new ones, which are the ones on which the test tells a 6 from an 8.

On letters under the root I always work with the same exercise in two versions, one with numbers and one with parameters: the absolute value is understood when you see it vanish in the numerical case and appear in the literal one. Lessons are online and one-to-one: for current rates see the pricing page, useful for assessing the 10-hour package if Year 2 has got off to a bad start and factorisation also needs putting right. After radicals, in Year 2, the absolute value that appears under the root returns in irrational inequalities (absolute value and irrational equations in Year 3) and calculations with irrational coefficients arrive in systems (linear systems: the four methods); and the first of the shortcuts that look right, $a^{2} = a$ , is explained in five maths mistakes that look right.

The quickest way to see where you stand is to try the radicals practice test: fifteen questions in fifty minutes, like the real test, with a breakdown of weak points by topic at the end. The correct answer to every question is free and needs no registration; the guided step-by-step solution unlocks with your first lesson. If after the test you see that the properties hold up but the conditions of existence don’t (or vice versa), write to me for a free introductory call: we look at the syllabus of your test and I tell you honestly how many lessons you need. The full picture of how I set up the path is on the page for maths tutoring.

Radicals and rationalisation: preparing the Year 2 test